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Govt Engineering college k r pete

Database Applications Laboratory

1

1. Consider the Insurance database given below. The primary keys are underlined and the

data types are specified:

PERSON (driver-id:string,name:string,address:string)

CAR(Regno:string,model:string,year:int)

ACCIDENT (report-number:int,date:date,location:string)OWNS (driver-id:string,regno:string)

PARTICIPATED (driver-id:string,regno:string,report-number:int,damage-amount:int)

i) create the above tables by properly specifying the primary keys and the foreign keys

ii) Enter atleast five tuples for each relation

iii) Demonstrate how you

a. Update the damage amount for the car with a specific regno in accident with

report number 12 to 25000

b. Add a new accident to the database

iv) Find the total number of people who owned cars that were involved in accidents in

2006.

v) Find the number of accidents in which cars belonging to a specific model were

involved.

vi) Generation of suitable reports

vii)Create suitable front end for querying and display the results

create table person

(

driverid varchar(5) primary key,

name varchar(30),

address varchar(50));

create table car

(

regno varchar(5) primary key,

model varchar(25),

year int

);

create table accident

(reportno int primary key,

date1 date,

location varchar(30)

);

create table owns

(

driverid references person,

regno references car

);


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create table participated

(

driverid references person,

regno references car,

reportno references accident,damages int

);

insert into person values (1,'abc','hkbkce,nagawara');

insert into person values (2,'bcd','atria,yelahanka');

insert into person values (3,'ade','ramahiah,near iisc');

insert into person values (4,'cfg','mvjce,whitefield');

insert into person values (5,'rte','pesit,somewhere');

insert into car values ('a1','merc',2007);

insert into car values ('a2','porsche',2005);

insert into car values ('b2','rolls royce',2006);

insert into car values ('b4','ferrari',1999);

insert into car values ('c1','bentley',2000);

insert into car values ('d2','maybach',2007);

insert into accident values (1,'1-jan-2006','domlur');

insert into accident values (2,'28-mar-2006','m.g.road');

insert into accident values (3,'2-dec-2006','r.t.nagar');

insert into accident values (4,'5-jan-2007','yelahanka');

insert into accident values (5,'26-jan-2007','koramangla');insert into accident values (6,'4-feb-2007','majestic');

insert into owns values (1,'a1');

insert into owns values (2,'a2');

insert into owns values (2,'c1');

insert into owns values (3,'b2');

insert into owns values (4,'b4');

insert into owns values (5,'d2');

insert into participated values (1,'a1',1,500);

insert into participated values (2,'a2',1,2000);insert into participated values (3,'b2',1,1000);


insert into participated values (4,'c1',2,800);

insert into participated values (5,'d2',3,750);

insert into participated values (2,'c1',3,600);


insert into participated values (5,'d2',4,1000);


insert into participated values (3,'b2',6,10000);



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select * from person;

DRIVERID NAME ADDRESS

---------------- ---------- ----------------------

1 abc hkbkce,nagawara

2 bcd atria,yelahanka

3 ade ramahiah,near iisc4 cfg mvjce,whitefield

5 rte pesit,somewhere

select * from car;

REGNO MODEL YEAR

----------- ---------------- ------------

a1 merc 2007

a2 porsche 2005

b2 rolls royce 2006

b4 ferrari 1999

c1 bentley 2000

d2 maybach 2007

select * from accident;

REPORTNO DATE1 LOCATION

----------------- -------------- -----------------------

1 01-JAN-06 domlur

2 28-MAR-06 m.g.road

3 02-DEC-06 r.t.nagar

4 05-JAN-07 yelahanka

5 26-JAN-07 koramangla6 04-FEB-07 majestic

select * from owns;

DRIVERID REGNO

-------------- -----------

1 a1

2 a2

2 c1

3 b2

4 b4

5 d2

select * from participated;

DRIVERID REGNO REPORTNO DAMAGES

---------------- ----------- ----------------- ----------------

1 a1 1 500

2 a2 1 2000

3 b2 1 1000

1 a1 2 1500

4 c1 2 800

5 d2 3 750

2 c1 3 6001 a1 3 200


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5 d2 4 1000

2 a2 5 1200

3 b2 6 10000

2 a1 6 5000

iii.a Update the damage amount for the car with a specific regno in accident with reportnumber 12 to 25000

update participated set damages=25000 where reportno=4 and regno='d2';

1 row updated



---------------- ----------- ----------------- ----------------

1 a1 1 500

2 a2 1 2000

3 b2 1 1000

1 a1 2 1500

4 c1 2 800

5 d2 3 750

2 c1 3 600

1 a1 3 200

5 d2 4 25000

2 a2 5 1200

3 b2 6 10000

2 a1 6 5000

iii.b Add a new accident to the database

insert into accident values (7,'16-mar-2007','nagawara');

1 row updated


1 row updated

select * from accident;

REPORTNO DATE1 LOCATION----------------- -------------- -----------------------

1 01-JAN-06 domlur

2 28-MAR-06 m.g.road

3 02-DEC-06 r.t.nagar

4 05-JAN-07 yelahanka

5 26-JAN-07 koramangla

6 04-FEB-07 majestic

7 16-MAR-07 nagawara


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---------------- ----------- ----------------- ----------------

1 a1 1 500

2 a2 1 2000

3 b2 1 10001 a1 2 1500

4 c1 2 800

5 d2 3 750

2 c1 3 600

1 a1 3 200

5 d2 4 25000

2 a2 5 1200

3 b2 6 10000

2 a1 6 5000

2 a1 7 15000

iv. Find the total number of people who owned cars that were involved in accidents in 2006

select count(distinct p.driverid) from accident a,owns o,participated p

where date1 like '%06' and a.reportno=p.reportno

and p.regno=o.regno and o.driverid=p.driverid;

COUNT(DISTINCTP.DRIVERID)

---------------------------------------------

4

v. Find the number of accidents in which cars belonging to a specific model were involved.select count(*) from car c,participated p where model='porsche' and c.regno=p.regno;

COUNT(*)

--------------

2


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2. Consider the following relations for an order processing database application in a

company.

CUSTOMER(Cust #: int, Cname: string, City: string)

ORDER(Order #: int, Odate: date, Cust #: int, Ord-Amt: int)

ORDER-ITEM (Order #: int, Item #: int, qty: int)ITEM (Item #: int, Unit Price: int)

SHIPMENT (Order #: int, Warehouse #: int, Ship-Date: date)

WAREHOUSE (Warehouse #: int, City: string)

i) Create the above tables by properly specifying the primary keys and the foreign keys.

ii) Enter at least five tuples for each relation.

iii) Produce a listing: CUSTNAME, NO_OF_ORDERS, AVG_ORDER_AMT, where

the middle column is the total number of orders by the customer and the last column

is the average order amount for that customer.

iv) List the Order# for the orders that were shipped from all the warehouses that the

company has in a specific city.

v) Demonstrate how you delete Item# 10 from the ITEM table and make that field nullin the ORDER-ITEM table.

vi) Generation of suitable reports.

vii) Create a suitable front end for querying and displaying the results.

create table customer

(

cno int primary key,

cname varchar(10),

city varchar(15));

create table order1

(

ono int primary key,

odate date,

cno references customer,

oamt int

);

create table item(

ino int primary key,

uprice int

);

create table oitem

(

ono references order1,

ino references item on delete set null,

qty int

);


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create table warehouse

(

wno int primary key,

city varchar(15)

);

create table shipment

(

ono references order1,

wno references warehouse,

sdate date

);

insert into customer values(1,'abc','bangalore');

insert into customer values(2,'cde','mysore');

insert into customer values(3,'def','chennai');

insert into customer values(4,'efg','mumbai');

insert into customer values(5,'adf','kolkata');

insert into order1 values(1,'1-jan-2006',1,20000);

insert into order1 values(2,'26-mar-2006',2,10000);

insert into order1 values(3,'12-jun-2006',1,5000);

insert into order1 values(4,'15-sep-2006',3,9000);



insert into order1 values(7,'3-mar-2007',5,3500);

insert into item values(1,500);





insert into oitem values(1,1,40);




insert into oitem values(4,2,30);insert into oitem values(6,4,3);


insert into warehouse values(100,'bangalore');

insert into warehouse values(101,'chennai');

insert into warehouse values(102,'mumbai');

insert into warehouse values(103,'kolkata');

insert into warehouse values(104,'mysore');

insert into shipment values(1,100,'3-jan-2006');

insert into shipment values(2,100,'28-mar-2006');insert into shipment values(3,101,'13-jun-2006');


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insert into shipment values(4,102,'18-sep-2006');



insert into shipment values(7,103,'3-mar-2007');

select * from customer;

CNO CNAME CITY

---------- ---------- ---------------

1 abc bangalore

2 cde mysore

3 def chennai

4 efg mumbai

5 adf kolkata

select * from order1;

ONO ODATE CNO OAMT

---------- ---------------- ---------- ----------

1 01-JAN-06 1 20000

2 26-MAR-06 2 10000

3 12-JUN-06 1 5000

4 15-SEP-06 3 9000

5 05-JAN-07 4 2500

6 10-JAN-07 4 2400

7 03-MAR-07 5 3500

select * from item;

INO UPRICE

---------- ----------

1 500

2 300

3 2500

4 800

5 700

select * from oitem;

ONO INO QTY

---------- ---------- ----------

1 1 40

2 1 20

3 3 2

5 3 1

4 2 30

6 4 3

7 5 5


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select * from warehouse;

WNO CITY

---------- ---------------

100 bangalore

101 chennai

102 mumbai103 kolkata

104 mysore

select * from shipment;

ONO WNO SDATE

---------- -------- ------------

1 100 03-JAN-06

2 100 28-MAR-06

3 101 13-JUN-06

4 102 18-SEP-06

5 103 11-JAN-07

6 104 13-JAN-07

7 103 03-MAR-07

iii. Produce a listing: CUSTNAME, NO_OF_ORDERS, AVG_ORDER_AMT, where the

middle column is the total number of orders by the customer and the last column is the

average order amount for that customer.

select c.cname,count(q.qty),avg(oamt)from customer c, order1 o, oitem q

where q.ono=o.ono and c.cno=o.cno

group by cname;

CNAME COUNT(Q.QTY) AVG(OAMT)

------------ ---------------------- --------------------

abc 2 12500

adf 1 3500

cde 1 10000

def 1 9000

efg 2 2450

iv. List the Order# for the orders that were shipped from all the warehouses that the

company has in a specific city.

select w.city, s.wno, s.ono

from warehouse w, shipment s

where w.wno=s.wno and w.city='mumbai';

CITY WNO ONO

--------------- ---------- ----------

mumbai 102 4


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v. Demonstrate how you delete Item# 10 from the ITEM table and make that field null in

the ORDER-ITEM table

delete from item where ino=5;

1 row deleted

select * from item;INO UPRICE

---------- ----------

1 500

2 300

3 2500

4 800


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3. Consider the following database of student enrollement in courses and books adopted for

each course .

STUDENT (regno :string , name : string , major : string , bdate : int)

COURSE (course# : int , cname : string , dept : string)

ENROLL ( regno : string , course#: int , sem : int , marks : int )BOOK_ADAPTION ( course#: int , sem : int , book_isbn :int)

TEXT( book_isbn : int , book-title : string , publisher : string , author : string).

i) Create the above tables by properly specifying the primary keys and the foreign key .

ii) Enter atleast five tuples for each relation .

iii)Demonstrate how you add a new text book to the database and make this book be

adopted by some department.

iv)Produce a list of text books( include course # ,book_isbn,book-title) in the

alphabetical order for courses offered by the cs department that use more than 2 books.

v) List any department that has all its adopted books published by specific publisher.

vi)Generation of suitable reports.

vii) Create suitable front end for querying and display the results

create table student

(

regno varchar(5) primary key,

name varchar(10),

major varchar(10),

bdate date

);

create table course

(

cno int primary key,

cname varchar(10),

dept varchar(10)

);

create table enroll

(

regno references student,

cno references course,sem int primary key,

marks int

);

create table text

(

isbn int primary key,

title varchar(15),

publisher varchar(10),

author varchar(10)

);


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create table bad

(

cno references course,

sem references enroll,

isbn references text

);

insert into student values('cs42','mikhil','cse','17-dec-1986');

insert into student values('cs48','mujeeb','cse','02-sep-1986');

insert into student values('ec26','pradeep','cse','16-aug-1987');

insert into student values('ee37','majid','ise','28-may-1986');

insert into student values('is48','wajid','ise','28-may-1986');

insert into course values(1,'.net','computer');

insert into course values(2,'j2ee','computer');

insert into course values(3,'mis','infosci');

insert into course values(4,'fs','infosci');

insert into course values(5,'oracle','computer');

insert into enroll values('cs42',1,6,98);

insert into enroll values('cs48',2,3,97);

insert into enroll values('ec26',5,5,50);

insert into enroll values('is48',3,7,90);

insert into enroll values('ee37',4,4,80);

insert into text values(101,'let us c','lpe','fahad');

insert into text values(102,'c++','abc','mujeeb');insert into text values(103,'oracle','def','othman');

insert into text values(104,'.net','lpe','naushad');

insert into text values(105,'j2ee','pearson','mikhil');

insert into bad values(1,6,101);





select * from student;REGNO NAME MAJOR BDATE

---------- ---------- ------------- -------------

cs42 mikhil cse 17-DEC-86

cs48 mujeeb cse 02-SEP-86

ec26 pradeep cse 16-AUG-87

ee37 majid ise 28-MAY-86

is48 wajid ise 28-MAY-86


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select * from course;

CNO CNAME DEPT

---------- -------------- ------------

1 .net computer

2 j2ee computer

3 mis infosci4 fs infosci

5 oracle computer

select * from enroll;

REGNO CNO SEM MARKS

------------ ---------- ---------- ------------

cs42 1 6 98

cs48 2 3 97

ec26 5 5 50

is48 3 7 90

ee37 4 4 80

select * from text;

ISBN TITLE PUBLISHER AUTHOR

---------- --------------- ------------------ ---------------

101 let us c lpe fahad

102 c++ abc mujeeb

103 oracle def othman

104 .net lpe naushad

105 j2ee pearson mikhil

select * from bad;

CNO SEM ISBN

---------- ---------- ----------

1 6 101

1 6 103

1 6 102

4 4 104

5 7 105

iii.Demonstrate how you add a new text book to the database and make this book be

adopted by some department.insert into text values(106,'JAVA','pearson','Avril');


select * from text;

ISBN TITLE PUBLISHER AUTHOR

---------- --------------- ------------------ ---------------

101 let us c lpe fahad

102 c++ abc mujeeb

103 oracle def othman

104 .net lpe naushad

105 j2ee pearson mikhil106 JAVA pearson Avril


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select * from bad;

CNO SEM ISBN

---------- ---------- ----------

1 6 101

1 6 103

1 6 1024 4 104

5 7 105

2 7 106

iv. Produce a list of text books( include course # ,book_isbn,book-title) in the

alphabetical order for courses offered by the cs department that use more than 2 books.

SELECT s.cno,t.title,b.isbn

FROM course s,bad b,text t

where s.cno=b.cno

AND b.isbn=t.isbn

AND s.dept='computer'

AND b.cno in (select b.cno

from bad b,course s

where b.cno=s.cno

group by b.cno

having count(*)>2)

order by t.title;

CNO TITLE ISBN

---------- --------------- ----------1 c++ 102

1 let us c 101

1 oracle 103

v. List any department that has all its adopted books published by specific publisher.

select s.dept, t.publisher

from course s, bad b, text t

where s.cno=b.cno

and t.isbn=b.isbn

and publisher='lpe';

DEPT PUBLISHER

------------- --------------------

computer lpe

infosci lpe


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4. Consider the following relations for the details maintained by a book dealer.

AUTHOR(Author-id: int, Name: string, City: string, Country: string)

PUBLISHER(Publisher-id: int, Name: string, City: string, Country: string)

CATALOG (Book-id: int, title: string, author-id: int, Publisher-id: int, Category-id: int, Year:

int, Price: int)CATEGORY (Category-id: int, Description: string)

ORDER-DETAILS (Order-no : int, Book-id: int, Quantity: int)

i. Create the above tables by properly specifying the primary keys and the foreign keys.

ii. Enter at least five tuples for each relation.

iii. Give the details of the authors who have 2 or more books in the catalog and the price of

the books is greater than the average price of the books in the catalog and the year of

publication is after 2000.

iv. Find the author of the book which has maximum sales.

v. Demonstrate how you increase the price of books published by a specific publisher by

10%.

vi. Generation of suitable reports.

vii. Create a suitable front end for querying and displaying the results.

create table author

(

aid int primary key,

name varchar(15),

city varchar(15),

country varchar(15)

);

create table publisher

(

pid int primary key,

name varchar(15) ,

city varchar(15) ,

country varchar(15)

);

create table category

(

cid int primary key,description varchar(15)

);

create table catalog

(

bid int primary key,

title varchar(15),

aid references author,

pid references publisher,

cid references category,

year int,price int );


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create table odetails

(

ono int,

bid references catalog,

qty int

);

insert into author values (1,'mik','bang','ind');

insert into author values (2,'muj','bang','ind');

insert into author values (3,'prad','tri','aus');

insert into author values (4,'maj','anan','ame');

insert into author values (5,'waj','anan','euro');

insert into publisher values (101,'pearson','bang','ind');

insert into publisher values (102,'tata','mumbai','aus');

insert into publisher values (103,'sapna','che','euro');

insert into publisher values (104,'abc','tri','ame');insert into publisher values (105,'xyz','anan','ind');

insert into category values (1001,'computer');

insert into category values (1002,'electronics');

insert into category values (1003,'maths');

insert into category values (1004,'science');

insert into category values (1005,'electrical');

insert into catalog values (111,'lib1',1,101,1001,2002,500);



insert into catalog values (114,'lib4',4,104,1001,2006,350);insert into catalog values (115,'lib5',5,105,1004,2007,100);




insert into odetails values (1,111,2);




insert into odetails values (5,114,2);insert into odetails values (6,115,1);



select * from author;

AID NAME CITY COUNTRY

---------- --------------- --------------- ---------------

1 mik bang ind

2 muj bang ind

3 prad tri aus

4 maj anan ame5 waj anan euro


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select * from publisher;

PID NAME CITY COUNTRY

---------- --------------- --------------- ---------------

101 pearson bang ind

102 tata mumbai aus

103 sapna che euro104 abc tri ame

105 xyz anan ind

select * from category;

CID DESCRIPTION

---------- ---------------------

1001 computer

1002 electronics

1003 maths

1004 science

1005 electrical

select * from catalog;

BID TITLE AID PID CID YEAR PRICE

------- ------------ --------- -------- -------- ---------- ----------

111 lib1 1 101 1001 2002 500

112 lib2 2 102 1002 2000 800

113 lib3 3 103 1003 2003 200

114 lib4 4 104 1001 2006 350

115 lib5 5 105 1004 2007 100

116 lib6 2 103 1005 2007 600117 lib7 2 105 1002 2007 450

118 lib8 1 101 1001 2002 500

select * from odetails;

ONO BID QTY

---------- ---------- ----------

1 111 2

2 112 3

3 111 5

4 113 1

5 114 26 115 1

1 114 2

2 113 2

iii. Give the detail of the author who have 2 or more books in the catalog and the price

of the book is greater than the average price of the book in the catalog and the year of

publication after 2000

select name,city,country

from author

where aid in (select aid from catalog where year>2000 and price>(select avg(price) fromcatalog)


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group by aid having count(*)>1);

NAME CITY COUNTRY

----------- --------------- ---------------

mik bang ind

muj bang ind

iv. Find the author of the book which has maximum sales

select a.aid, name from author a, catalog c

where a.aid=c.aid and c.bid=(select bid from odetails group by bid having

sum(qty)=(select max(sum(qty)) from odetails group by bid));

AID NAME

---------- ---------------

1 mik

v. Demonstrate how to increase the price of books published by a specific publisher by

10%

update catalog set price=1.1*price where pid in (select pid from publisher where name='abc');

select * from catalog;

BID TITLE AID PID CID YEAR PRICE

------- ------------ --------- -------- -------- ---------- ----------

111 lib1 1 101 1001 2002 500

112 lib2 2 102 1002 2000 800

113 lib3 3 103 1003 2003 200

114 lib4 4 104 1001 2006 385

115 lib5 5 105 1004 2007 100116 lib6 2 103 1005 2007 600

117 lib7 2 105 1002 2007 450

118 lib8 1 101 1001 2002 500


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5. Consider the following database for a banking enterprise

BRANCH (branch_name: string, branch_city: string, assets: real)

ACCOUNT (accno: int, branch_name: string, balance: real)

CUSTOMER(customer_name: string, customer_street: string, city:string)DEPOSITOR(customer_name: string, accno: int)

LOAN (loan_number: int, branch_name: string, amount: real)

BORROWER(customer_name: string, loan_number: int)

i) Create the above tables by properly specifying the primary keys and the foreign keys.

ii) Enter atleast five tuples for each relation.

iii) Find all the customers who atleast two accounts at the MAIN branch.

iv) Find all the customers who have an account at all branches located in a specific city.

v) Demonstrate how you delete all account tuples at every branch located in a specific

city.

vi) Generation of suitable reports.

vii)Create suitable front end for querying and displaying the results.

create table branch

(

bname varchar(10) primary key,

city varchar(10),

assets real

);

create table account(

accno int primary key,

bname varchar(10),

balance real,

foreign key(bname) references branch(bname)

);

create table cust

(

cname varchar(10) primary key,

cstreet varchar(10) not null,city varchar(10) not null

);

create table depositor

(

cname varchar(10),

accno int,

primary key(cname,accno),

foreign key(accno) references account(accno) on delete cascade,

foreign key(cname) references cust(cname) on delete cascade

);


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create table loan

(

lno int primary key,

bname varchar(10) not null,

amt real,foreign key(bname) references branch(bname) on delete cascade

);

create table borrower

(

cname varchar(10),

lno int,

primary key(cname,lno),

foreign key(cname) references cust(cname) on delete cascade,

foreign key(lno) references loan(lno) on delete cascade

);

insert into branch values('abc','bang',1200000);

insert into branch values('def','che',2000000);

insert into branch values('abn','mum',330000);

insert into branch values('xyz','hyd',555555);

insert into branch values('mno','bang',9999999);

insert into account values(1,'abc',25000);

insert into account values(2,'def',12000);

insert into account values(3,'def',1000);insert into account values(4,'abn',10000);

insert into account values(5,'mno',600000);

insert into account values(6,'xyz',50000);

insert into cust values('mik','ab','bang');

insert into cust values('muj','cd','bang');

insert into cust values('maj','ef','che');

insert into cust values('waj','xy','del');

insert into cust values('prad','lm','mum');

insert into cust values('now','op','hyd');

insert into depositor values('mik',2);

insert into depositor values('muj',1);

insert into depositor values('muj',5);

insert into depositor values('prad',4);

insert into depositor values('maj',3);

insert into depositor values('waj',6);

insert into depositor values('mik',3);

insert into loan values(1,'abc',5000);

insert into loan values(2,'def',1500);

insert into loan values(3,'abn',10000);insert into loan values(4,'xyz',3500);


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insert into loan values(5,'mno',20000);

insert into borrower values('mik',2);

insert into borrower values('muj',1);

insert into borrower values('prad',3);

insert into borrower values('maj',4);insert into borrower values('waj',5);

select *from branch;

BNAME CITY ASSETS

---------- ---------- ----------

abc bang 1200000

def che 2000000

abn mum 330000

xyz hyd 555555

mno bang 9999999

select *from account;

ACCNO BNAME BALANCE

---------- ---------- ----------

1 abc 25000

2 def 12000

3 def 1000

4 abn 10000

5 mno 6000006 xyz 50000

select *from cust;

CNAME CSTREET CITY

---------- ------------ ----------

mik ab bang

muj cd bang

maj ef che

waj xy del

prad lm mumnow op hyd

select *from depositor;

CNAME ACCNO

---------- ------------

mik 2

muj 1

muj 5

prad 4

maj 3waj 6


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mik 3

select *from loan;

LNO BNAME AMT

---------- ------------ ----------

1 abc 5000

2 def 15003 abn 10000

4 xyz 3500

5 mno 20000

select *from borrower;

CNAME LNO

------------- ----------

mik 2

muj 1

prad 3

maj 4

waj 5

iii. Find all the customers who atleast two accounts at the MAIN branch.

select cname from account a,depositor d where a.accno=d.accno and bname='def' group by

cname having count(*)>1;

CNAME

----------

mik

iv. Find all the customers who have an account at all branches located in a specific cityselect cname from cust c where not exists

(select bname from branch where city='bang' minus select bname from depositor d,account a

where d.accno=a.accno and d.cname=c.cname)

and exists

(select bname from branch where city='bang');

CNAME

------------

muj

v. Demonstrate how you delete all account tuples at every branch located in a specific

city.delete from account where bname in (select bname from branch where city='che');

2 rows deleted.

select *from account;

ACCNO BNAME BALANCE

----------- ------------ ---------------

1 abc 25000

4 abn 10000

5 mno 600000

6 xyz 50000


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